Trying to show that columns of Fourier matrix are eigenvectors

Question

I posted a picture since the syntax for this one seems quite complex:

I found this: Discrete Fourier Transform - proof that columns of matrix are orthogonal which only shows that are orthogonal. In my case I need to show that they are eigenvectors. I have no idea how show this

If you have a matrix $A$ and a vector $x$, how do you show that $x$ is an eigenvector of $A$? Easy: just calculate $Ax$, and show that the result is a multiple of $x$. — Ben Grossmann, Apr 16 '17 at 15:49
Have you at least tried to do that much? If so, where are you getting stuck? — Ben Grossmann, Apr 16 '17 at 15:55
So you're saying to pick a random vector $x$, multiply it by A and then show the result is a multiple of A? — john, Apr 16 '17 at 15:58
Not a random vector: take $x$ to be a column vector from $F_n$. — Ben Grossmann, Apr 16 '17 at 16:06
@john Hi john, Did you find a simpler answer to this question? It's important to me, I would be so grateful if you help me with it. — user137927, Sep 19 '20 at 07:28

score 2 · Answer 1 · answered Apr 16 '17 at 15:58

Let $S$ be the cyclic shift operator. Notice that $A = -S^{-1} +2S^0 - S$. The fact that the discrete Fourier basis vectors are eigenvectors of $A$ now follows from the fact that they are eigenvectors of $S$. (And the whole point of the discrete Fourier basis vectors is that they are eigenvectors of $S$.)

A similar argument shows that the discrete Fourier basis vectors are eigenvectors for any convolution operator.

Trying to show that columns of Fourier matrix are eigenvectors

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